What is (\partial)/(\partial x)(2x^2-x^2y^3+y^2) ?

The answer to (\partial)/(\partial x)(2x^2-x^2y^3+y^2) is 4x-2xy^3

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(\partial)/(\partial x)(2x^2-x^2y^3+y^2)

\frac{\partial}{\partial x} (2 x^{2} - x^{2} y^{3} + y^{2})

4 x - 2 x y^{3}

Solution steps

\frac{\partial}{\partial x} (2 x^{2} - x^{2} y^{3} + y^{2})

Treat

y

as a constant

Apply the Sum/Difference Rule:

(f \pm g)^{'} = f^{'} \pm g^{'}

= \frac{\partial}{\partial x} (2 x^{2}) - \frac{\partial}{\partial x} (x^{2} y^{3}) + \frac{\partial}{\partial x} (y^{2})

\frac{\partial}{\partial x} (2 x^{2}) = 4 x

\frac{\partial}{\partial x} (x^{2} y^{3}) = 2 y^{3} x

\frac{\partial}{\partial x} (y^{2}) = 0

= 4 x - 2 y^{3} x + 0

Simplify

= 4 x - 2 x y^{3}

\int_{- 1}^{1} \frac{e ^{x}}{e ^{x} - 1} d x

\frac{d}{d x} (ln (\frac{2 x + 3}{2 x - 3}))

\frac{d}{d x} (e^{x^{3} + 5})

\int x^{- 4} d x

\frac{x ^{2} + 1}{x ^{3} - 1}

What is (\partial)/(\partial x)(2x^2-x^2y^3+y^2) ?
The answer to (\partial)/(\partial x)(2x^2-x^2y^3+y^2) is 4x-2xy^3